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3.2 moles of HI ( g ) were heated in a sealed bulb at 444 ° C till the equilibrium was reached . It's degree of dissociation was found to be 20%. Calculate the number of moles of hydrogen iodide, hydrogen and iodine present at the equilibrium point and determine the value of equilibrium constant for the reaction 2HI (g) (g). Considering the volume of the container 1 L. 2 (g) + I 2 H ( Please explain in steps.)
3.2 moles of HI(g) were heated in a sealed bulb at 444°C till the equilibrium was reached. It's degree of dissociation was found to be 20%. Calculate the number of moles of hydrogen iodide, hydrogen and iodine present at the equilibrium point and determine the value of equilibrium constant for the reaction  2HI (g)  (g). Considering the volume of the container 1 L.2(g) + I2  H (Please explain in steps.)

```
3 years ago

Arun
25768 Points
```							Concentration of HI is 3.2moles/1litrre=3.2Degree of dissociation=20%Degree of dissociation=no.of moles of reactant dissociated/no.of moles of reactants present initially 20/100=x3/2x=0.64molesAccording to the reaction2Hl(g)=H2(g)+l2(g)Two M of Hl gives one M of H2 & l2 each so 64M of HI would give 32M OF H2 & I2 each hence equilibrium concentration of HI is 2.56MKc=(0.32)(0.32)/(2.56)2=0.0156
```
3 years ago
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